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Chapter 6

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Stability

- A system is stable if the natural response approaches zero as time approaches infinity
- A system is unstable if the natural response approaches infinity as time approaches infinity
- A system is marginally stable if the natural response neither decays nor grows but remains constant or oscillates.
- A system is stable if every bounded input yields a bounded output
- A system is unstable if any bounded input yields an unbounded output

Stable systems have closed-loop transfer functions with poles in the left half-plane.Unstable systems have closed-loop transfer functions with at least one pole in the right half-plane, and/or poles of multiplicity greater than one on the imaginary axis Marginally stable systems have closed-loop transfer functions with only imaginary axis poles of multiplicity one and poles in the left half-plane.

Common cause of problems in finding closed-loop poles:

a. original system;b. equivalent system

Using this method we can tell how many closed-loop poles are in the left half-plane, in the right half-plane and on the imaginary axis.

The method requires two steps: (1) Generate the data table (Routh table) and (2) Interpret the table to determine the number of poles in LHP and RHP.

Interpretation of Routh table

The number of roots of the polynomial that are in the right half-plane is equal to the number of sign changes in the first column.

Replace the zero with , the value of is then allowed to approach zero from –ive or +ive side.

ProblemDetermine the stability of the closed-loop transfer function

R-H: Special case. Zero in the first column

ProblemDetermine the stability of the closed-loop transfer function

SolutionWrite a polynomial that has the reciprocal roots of the denominator

R-H: Special case. Entire row is zero

ProblemDetermine the number of right-half-plane poles in the closed transfer function

Solution: Form an auxiliary polynomial, P(s) using the entries of row above row of zeros as coefficient, then differentiate with respect to s finally use coefficients to replace the rows of zeros and continue the RH procedure.

An entire row of zeros will appear when a purely even or odd polynomial is a factor of the original polynomial.

Even polynomials have roots that are symmetrical about the origin.

If we don’t have a row of zeros, we cannot possibly have roots on the imaginary axis.

ProblemDetermine the number of poles in the right-half-plane, left-half-plan and on the axis for the closed transfer function

ProblemDetermine the number of poles in the right-half-plane, left-half-plan and on the axis for system

Solution: The closed loop transfer function is

2 poles in RHP, 2 poles in LHP no poles on axis The system is unstable

ProblemDetermine the number of poles in the right-half-plane, left-half-plan and on the axis for system

Solution: The closed loop transfer function is

Table for polynomial whose roots are the reciprocal of the original

2 sign changes, 2 poles in RHP system is unstable

ProblemDetermine the number of poles in the right-half-plane, left-half-plan and on the axis for system. Draw conclusions about stability of the closed loop system.

Solution: The closed loop transfer function is

There is a row of zeros in the s5, so the s6 row forms an even polynomial. 2 sign changes from the even polynomial so 2 poles in the RHP and because of symmetry about origin 2 will be in the LHP

The 2 remaining poles are on the axis

ProblemFind the range of gain K for the system that will cuase the system to be stable, unstabel, and marginally stable. Assume K>0.

Solution: The closed loop transfer function is

For K < 1386 the system is stable. For K > 1386 the system is unstable. For K = 1386 we will have entire row of zeros (s row). We form the even polynomial and differentiate and continue, no sign changes from the even polynomial so the 2 roots are on the axis and the system is marginally stable

Problem Factor the polynomial

Solution: from the Routh table we see that the s1 row is a row of zeros. So the even polynomial at the s2 row is since this polynomial is a factor of the original, dividing yields

As the other factor so

Problem Given the system

Find out how many poles in the LHP, RHP and on the axis

Solution: First form (sI-A)

Now find the det (sI-A) =

Routh table for Example 6.11

One sign change, so 1 pole in the LHP and the system is unstable